Question

A voltaic cell is constructed that uses the following halfcell reactions: $$ \begin{gathered} \mathrm{Cu}^{+}(a q)+\mathrm{e}^{-} \longrightarrow \mathrm{Cu}(s) \\ \mathrm{I}_{2}(s)+2 \mathrm{e}^{-} \longrightarrow 2 \mathrm{I}^{-}(a q) \end{gathered} $$ The cell is operated at \(298 \mathrm{~K}\) with \(\left[\mathrm{Cu}^{+}\right]=0.25 \mathrm{M}\) and \(\left[\mathrm{I}^{-}\right]=3.5 \mathrm{M}\). (a) Determine \(E\) for the cell at these concentrations. (b) Which electrode is the anode of the cell? (c) Is the answer to part (b) the same as it would be if the cell were operated under standard conditions? (d) If \(\left[\mathrm{Cu}^{+}\right]\) was equal to \(0.15 \mathrm{M}\), at what concentration of \(\mathrm{I}^{-}\) would the cell have zero potential?

Short answer

(a) The cell potential at the given concentrations is 0.014 V. (b) The anode of the cell is the electrode where the Cu⁺/Cu half-cell reaction takes place. (c) Yes, the anode and cathode will remain the same under standard conditions. (d) At [Cu⁺] = 0.15 M, the concentration of I⁻ required for the cell to have zero potential is approximately 0.048 M.

Step-by-step solution

Find the standard reduction potentials

The standard reduction potentials for the half-cell reactions are: \(E^\circ_{\mathrm{Cu}^+/Cu}=+0.52\,\,\text{V}, \,\, E^\circ_{I_2/I^{-}}=+0.54\,\,\text{V}\) Note that these values are given in the textbooks or tables. Now we can find the standard cell potential by subtracting the standard reduction potential of the half-cell reaction that occurs at the anode (oxidation) from the potential of the reaction that occurs at the cathode (reduction).

Determine the anode and cathode

Since \(E^\circ_{I_2/I^{-}} > E^\circ_{\mathrm{Cu}^+/Cu}\), Iodine will be reduced (gain electrons) to form I⁻, and Cu⁺ will be oxidized (lose electrons) to form Cu. Therefore, Cu⁺/Cu half-cell will be the anode.

Calculate the standard cell potential

Now let's calculate the standard cell potential: \(E^\circ_\text{cell} = E^\circ_{I_2/I^{-}} - E^\circ_{\mathrm{Cu}^+/Cu} = 0.54 - 0.52 = 0.02\,\,\text{V}\)

Calculate the cell potential

With the standard cell potential and the given concentrations, we can now apply the Nernst equation to find the cell potential: \(E_\text{cell} = E^\circ_\text{cell} - \frac{RT}{nF} \ln Q\) where: - \(R\) is the gas constant (8.314 J/(mol.K)), - \(T\) is the temperature (298 K), - \(n\) is the number of electrons transferred (2), - \(F\) is the Faraday constant (96,485 C/mol), and - \(Q\) is the reaction quotient, which is equal to \(\frac{[\mathrm{Cu}^+]}{[\mathrm{I}^{-}]^2}\). Plug the given values into the equation: \(E_\text{cell} = 0.02 - \frac{(8.314)(298)}{(2)(96,485)}\ln \frac{(0.25)}{(3.5)^2}\) Now, by calculating all the values we get: \(E_\text{cell} = 0.02 -\left(-\frac{1}{2}\right)*(-0.026 \,\,\text{V}\) = 0.014 \,\text{V}\) (a) The cell potential at the given concentrations is 0.014 V. (b) The anode of the cell is the electrode where the Cu⁺/Cu half-cell reaction takes place. (c) Yes, the anode and cathode will remain the same under standard conditions. (d) If [Cu⁺] = 0.15 M and E_cell = 0, we can solve for the I⁻ concentration using the Nernst equation: \(0=0.02- \frac{((8.314)(298)}{(2)(96,485)}\ln \frac{(0.15)}{[\mathrm{I}^{-}]^2}\) Rearrange and solve for [I⁻]: \(\ln\frac{0.15}{[\mathrm{I}^{-}]^2}=\frac{0.02*(2)*(96,485)}{(8.314)(298)}\) Let x = [I⁻], and after calculating the numerical value and taking the exponential: \(\frac{0.15}{x^2} = e^{12.98}\) Rearrange and solve for x: \(x^2= \frac{0.15}{e^{12.98}}\) \(x=[\mathrm{I}^{-}] = \frac{\sqrt{(0.15)}}{\sqrt{(e^{12.98})}}\approx 0.048 \,\,\text{M}\) At [Cu⁺] = 0.15 M, the concentration of I⁻ required for the cell to have zero potential is approximately 0.048 M.

Key concepts

Standard Reduction Potentials

Understanding the standard reduction potentials is critical in electrochemistry, particularly when working with voltaic cells. These potentials, represented by the symbol \( E^\circ \), measure the tendency of a chemical species to be reduced, that is, to gain electrons. Values are given under standard conditions, which means a temperature of 298 K, a pressure of 1 atm, and solute concentrations at 1 M.

In a voltaic cell, every half-cell has its own standard reduction potential. By comparing these values, we can determine which substance will act as the oxidizing agent (the one being reduced) and which will be the reducing agent (the one being oxidized). For instance, in our example with copper and iodine, copper has a lower standard reduction potential than iodine; thus, it more readily loses electrons, serving as the anode. On the contrary, iodine has a higher potential and gains electrons, functioning as the cathode. It's essential to look up these potentials in a standard reference table to solve problems related to cell potential calculations.

Nernst Equation

The Nernst equation is fundamental to electrochemistry, providing the means to calculate the potential of a voltaic cell under non-standard conditions. This equation allows us to incorporate the effect of different temperatures, pressures, and concentration of ions into the calculation of cell potential. The equation is stated as:\( E = E^\circ - \frac{RT}{nF} \ln Q \)
where \( E \) is the cell potential at non-standard conditions, \( E^\circ \) is the standard cell potential, \( R \) is the gas constant, \( T \) is the temperature in Kelvin, \( n \) is the number of moles of electrons exchanged in the half-cell reactions, \( F \) is the Faraday constant, and \( Q \) is the reaction quotient. The reaction quotient represents the ratio of products over reactants, taking into account their concentrations or partial pressures.

When dealing with the Nernst equation, it's vital to express all units correctly and to understand that as the concentrations of the reactants increase or the products decrease, the cell potential will decrease from its standard value, as observed in the example problem.

Cell Potential Calculation

Calculating the cell potential of a voltaic cell is a sequential process. Initially, you determine the standard cell potential by finding the difference between the standard reduction potentials of the cathode and anode. The formula looks like this:\( E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \)
Once the standard potential is known, we can consider the actual conditions inside the cell, which might differ from standard conditions. If concentrations, temperature, or pressure are not at standard state, we use the Nernst equation, as mentioned earlier, to adjust our calculation. This allows us to predict how the cell will behave under various internal and external conditions, giving us a deeper insight into its potential at any given moment.

Half-Cell Reactions

Voltaic cells operate based on half-cell reactions, consisting of an oxidation half-reaction and a reduction half-reaction. In these reactions, electrons are transferred from one species to another. The site where oxidation takes place is called the anode and reduction occurs at the cathode.

In-depth comprehension of half-cell reactions is essential since they tell us not only which species are being oxidized or reduced but also how many electrons are involved in these reactions. This knowledge is necessary to balance chemical equations, calculate cell potentials, and understand the function of the voltaic cell.

The half-reaction with the higher standard reduction potential will go forward as the reduction reaction at the cathode whereas the half-reaction with the lower potential will be reversed to proceed as the oxidation reaction at the anode. Understanding and writing these half-cell reactions accurately sets the foundation for predicting the behavior and efficiency of the voltaic cell in question.